原题链接在这里：https://leetcode.com/problems/valid-anagram/
题目：
Given two strings s and t, write a function to determine if t is an anagram of s.
For example,
s = "anagram", t = "nagaram", return true.
s = "rat", t = "car", return false.
Note:
You may assume the string contains only lowercase alphabets.
Follow up:
What if the inputs contain unicode characters? How would you adapt your solution to such case?
题解：
维护一个map, s对应char位++, t对应char位--. 若是map有非零位便return false.
Time Complexity: O(n), n 为string长度.
Space: O(1). 用了一个array当map.
AC Java：
Method 1:
1 public class Solution { 2 public boolean isAnagram(String s, String t) { 3 if(s == null && t == null){ 4 return true; 5 } 6 if(s == null || t == null){ 7 return false; 8 } 9 if(s.length() != t.length()){ 10 return false; 11 } 12 13 int [] map = new int[256]; 14 for(int i = 0; i<s.length(); i++){ 15 map[s.charAt(i)]++; 16 } 17 for(int i = 0; i<t.length(); i++){ 18 map[t.charAt(i)]--; 19 } 20 for(int i = 0; i<256; i++){ 21 if(map[i] != 0){ 22 return false; 23 } 24 } 25 return true; 26 } 27 }
这道题可以直接sort 两个string 然后比较，相同就返回true.
不过要注意for Strings, equals() method works as str1.equals(str2). for Arrays, equals() method works as Arrays.equals(arr1, arr2).
Time Compleixty: O(nlogn). 有sort在内.
Space: O(n), string换成了char array.
Method 2:
1 public class Solution { 2 public boolean isAnagram(String s, String t) { 3 //Method 2 4 char[] temp1 = s.toCharArray(); 5 char[] temp2 = t.toCharArray(); 6 Arrays.sort(temp1); 7 Arrays.sort(temp2); 8 9 return Arrays.equals(temp1,temp2); 10 } 11 }